Question: Suppose that $80\%$ of trees in a forest of $300$ trees are infected with a virus, and we take an SRS of $5$ trees to test for the virus. What is the probability that exactly $2$ of the $5$ trees have the virus? You may round your answer to the nearest hundredth. $P(X=2)=$
Answer: Without a fancy calculator Getting a sample where $2$ of the $5$ trees have the virus means we need to get $2$ trees with the virus and $3$ trees that don't have the virus. We know $P({\text{virus}})={80\%}$ and $P({\text{no virus}})={20\%}$. We can assume independence since we are sampling less than $10\%$ of the population. So let's multiply probabilities to find the probability of getting $2$ trees with the virus followed by $3$ trees that don't have the virus: $P({\text{VV}}{\text{NNN}})=({0.8})^2({0.2})^3=0.00512$ This isn't our final answer, because there are other ways for $2$ of the $5$ trees to have the virus (for example, NNNVV). How many different ways are there? We can use the combination formula to find how many ways there are for $2$ of the $5$ trees to have the virus: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _5\text{C}_2&=\dfrac{5!}{(5-2)!\cdot2!} \\\\ &=\dfrac{5 \cdot 4 \cdot \cancel{3 \cdot 2 \cdot 1}}{(\cancel{3 \cdot 2 \cdot 1}) \cdot 2 \cdot 1} \\\\ &=10 \end{aligned}$ There are $10$ ways for $2$ of the $5$ trees to have the virus. Do they all have the same probability? Each of the $10$ ways has the same probability that we already found: $\begin{aligned} P({\text{VV}}{\text{NNN}})&=({0.8})^2({0.2})^3=0.00512 \\\\ P({\text{V}}{\text{N}}{\text{V}}{\text{NN}})&=({0.8})^2({0.2})^3=0.00512 \\\\ \vdots \\\\ P({\text{NNN}}{\text{VV}})&=({0.8})^2({0.2})^3=0.00512 \end{aligned}$ So we can multiply this probability by $10$ since that is how many ways there are for $2$ of the $5$ trees to have the virus: $\begin{aligned} P(X=2)&=10(0.8)^2(0.2)^3 \\\\ &=10(0.00512) \\\\ &=0.0512 \end{aligned}$ Answer $P(X=2)=0.0512\approx0.05$